Derive α and β for lossy dielectric

For a medium with ε, μ and σ, Maxwell's equations in phasor form give the wave equation ∇²E = γ²E, where

γ² = jωμ(σ + jωε)

(σ + jωε is the total conduction plus displacement current term.) Put γ = α + jβ.

Step 1: Expand γ²

γ² = (α + jβ)² = α² − β² + j2αβ
γ² = jωμσ − ω²με

Equate real and imaginary parts:

(1)  α² − β² = −ω²με
(2)  2αβ     = ωμσ

Step 2: Use the magnitude of γ²

|γ²| = α² + β² = ωμ √(σ² + ω²ε²)       (3)

Taking (3) and (1) together:

Adding (3) and (1):    2α² = ωμ√(σ² + ω²ε²) − ω²με
Subtracting (1) from (3): 2β² = ωμ√(σ² + ω²ε²) + ω²με

Step 3: Final expressions

α = ω √(με/2) · [ √(1 + (σ/ωε)²) − 1 ]^(1/2)   Np/m

β = ω √(με/2) · [ √(1 + (σ/ωε)²) + 1 ]^(1/2)   rad/m

Here σ/ωε = tan δ is the loss tangent.

Special cases

• Lossless dielectric (σ = 0): α = 0, β = ω√(με).
• Low-loss (σ ≪ ωε): α ≈ (σ/2)√(μ/ε), β ≈ ω√(με).
• Good conductor (σ ≫ ωε): α = β ≈ √(πfμσ).